SOLUTION: Hi there, I have to find a degree 3 polynomial whose coefficients are real numbers and with given zeros: {{{-3}}}, {{{1-2i}}} If {{{1-2i}}} is a zero, then so is {{{1+2i}}} so t

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hi there, I have to find a degree 3 polynomial whose coefficients are real numbers and with given zeros: {{{-3}}}, {{{1-2i}}} If {{{1-2i}}} is a zero, then so is {{{1+2i}}} so t      Log On


   

Question 365323: Hi there, I have to find a degree 3 polynomial whose coefficients are real numbers and with given zeros: -3, 1-2i
If 1-2i is a zero, then so is 1%2B2i so then -3%281-2i%29%281%2B2i%29=+0
Then using another example, I came up with this equation. But I'll be honest I don't understand why/how I did this??
%28x-1%2B2i%29%28x-1-2i%29%28x%2B3%29
Then solving through...
x%5E2-x-2ix-x%2B1%2B2ix-2i%2B4i%5E2
%28x%5E2-2x%2B5%29%28x%2B3%29
x%5E3%2B3x%5E2-2x%5E2-6x%2B5x%2B15
which gives me my answer x%5E3%2Bx%5E2-x%2B15, did I do this correctly?
Thanks so much for your time :)

Found 2 solutions by robertb, stanbon:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
You got it right. If you try to divide x%5E3%2Bx%5E2-x%2B15 by x+3 by synthetic division, you get the quotient x%5E2-2x%2B5+=+0 (a quadratic equation!) which has roots 1-2i and 1%2B2i (using the quadratic formula.)

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find a degree 3 polynomial whose coefficients are real numbers and with given zeros: -3, 1-2i
If 1-2i is a zero, then so is 1%2B2i so then -3%281-2i%29%281%2B2i%29=+0
Comment: That is not true; it is not what "zeroes" mean.
------------------------------
Then using another example, I came up with this equation. But I'll be honest I don't understand why/how I did this??
Comment: The rule is "If a is a zero, (x-a) is a factor"
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f(x) = (x-(1+2i))(x-(1-2i))(x--3)
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Rearrange to get:
f(x) = ((x-1)-2i)((x-1)+2i)(x+3)
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f(x) = [(x-1)^2 + 4][x+3]
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f(x) = [x^2-2x+5][x+3]
---
f(x) = (x^2-2x+5)x + (x^2-2x+5)3
---
f(x) = x^3-2x^2+5x+3x^2-6x+15
---
f(x) = x^3+x^2-x+15
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You were correct.
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Cheers,
Stan H.